Respuesta :
Answer:
(a). The potential is 78.0 kV.
(b). The potential is zero.
Explanation:
Given that,
Radius = 6.00
Charge density = 8.50 μC/m
(a). The surface of the cylinder and 4.00 cm away from the surface,
We need to calculate the voltage
Using formula of potential
[tex]V=\dfrac{\lambda}{2\pi\epsilon_{0}}ln(\dfrac{r}{R})[/tex]
Put the value into the formula
[tex]V=\dfrac{8.5\times10^{-6}}{2\pi\times8.85\times10^{-12}}ln(\dfrac{6+4}{6})[/tex]
[tex]V=78.0\ kV[/tex]
(b). The surface and a point 1.00 cm from the central axis of the cylinder
Here, r = R
We need to calculate the voltage
Using formula of potential
[tex]V=\dfrac{\lambda}{2\pi\epsilon_{0}}ln(\dfrac{r}{R})[/tex]
Put the value into the formula
[tex]V=\dfrac{8.5\times10^{-6}}{2\pi\times8.85\times10^{-12}}ln(1)[/tex]
[tex]V=0\ kV[/tex]
Hence, (a). The potential is 78.0 kV.
(b). The potential is zero.
